Question: What is the value of the following logarithm? $\log_{9} \left(\dfrac{1}{729}\right)$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $9^{y} = \dfrac{1}{729}$ In this case, $9^{-3} = \dfrac{1}{729}$, so $\log_{9} \left(\dfrac{1}{729}\right) = -3$.